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2 years ago

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Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp

ressed in g's where g = 9.8 m/s^2 1 point the same as in English units smaller larger this calculation cannot be performed in SI units If the initial velocity of player is doubled then the absolute acceleration would * 1 point decrease 2x stays the same increase 2x increase 4x If stopping distance is decreased 2x then the acceleration would * 1 point decrease 2x stay the same increase 2x increase 4x Other: Hockey player having initial velocity 20 m/s and helmet having padding 5 cm hits head on his twin brother. The acceleration of his brain is most closely equal to * 1 point 0.4g 4g 40g 400g

1 answer:

Elanso [62]2 years ago

6 0

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

v² = v₀² + 2 a x

0 = v₀² + 2 a x

a = - v₀² / 2 x

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

a/g = - (2v₀) 2 / 2xg

a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

a/g = (- v₀² / 2g) 2/x

a =2 (-v₀² / 2x₀g) g

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

a / g = - 20² / (2 0.05)

a / g = - 4000 / g

a / g = - 4000 / 9.8 = 408

a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

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Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

6 0

1 year ago

Two sinusoidal waves are identical except for their phase. When these two waves travel along the same string, for which phase di

Kamila [148]

Answer:

zero or 2π is maximum

Explanation:

Sine waves can be written

x₁ = A sin (kx -wt + φ₁)

x₂ = A sin (kx- wt + φ₂)

When the wave travels in the same direction

Xt = x₁ + x₂

Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]

We are going to develop trigonometric functions, let's call

a = kx + wt

Xt = A [sin (a + φ₁) + sin (a + φ₂)

We develop breasts of double angles

sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a

sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a

Let's make the sum

sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)

to have a maximum of the sine function, the cosine of fi must be maximum

cos φ₁ + cos φ₂ = 1 +1 = 2

the possible values of each phase are

φ1 = 0, π, 2π

φ2 = 0, π, 2π,

so that the phase difference of being zero or 2π is maximum

6 0

2 years ago

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

$2 n t = (m + \dfrac{1}{2})\lambda$

At m = 0 for the minimum thickness, so

$2\times 1.45 \times t = (0 + 0.5)\times 600$

$2.9 \times t =300$

$t =\dfrac{300}{2.9}$

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0

2 years ago

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

shutvik [7]

Answer: 9938.8 km

Explanation:

1 pound-force = 4.48 N

30.0 pounds-force = 134.4 N

The force of gravitation between Earth and object on the surface of is given by:

$F = \frac{GMm}{R^2} = mg$

Where M is the mass of the Earth, m is the mass of the object, R (6371 km) is the radius of the Earth.

At height, h above the surface of the Earth, the weight of the object:

$(mg)'= \frac{GMm}{(R+h)^2}$

we need to find "h"

taking the ratio of two:

$\frac{mg}{(mg)'}=\frac{(R+h)^2}{R^2}\\ \Rightarrow \frac{90kg \times 9.8 m/s^2}{134.4 N}=\frac{(R+h)^2}{R^2}\\ \Rightarrow 6.56 R^2= (R+h)^2 \Rightarrow h= (2.56-1)R\\ \Rightarrow h = 1.56 R = 1.56 \times 6371 km = 9938. 8 km$

Hence, Pete would weigh 30 pounds at 9938.8 km above the surface of the Earth.

5 0

2 years ago

A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

tino4ka555 [31]

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

$v = 70 m/s$

now we have

$v_x = 70 cos25$

$v_x = 63.44 m/s$

$v_y = 70 sin25$

$v_y = 29.6 m/s$

now in Y direction we can use kinematics

$v_y = v_i + at$

$v_y = 29.6 - (9.81 \times 5)$

$v_y = -19.5 m/s$

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{63.44^2 + 19.5^2}$

$v = 66.4 m/s$

4 0

1 year ago